What is solution of the differential equation:

\begin{aligned}x({yy'' + y'^2}) + yy' = 0\end{aligned}

What i am confused about it is the treatment of $$\left(\dfrac{dy}{dx}\right)^2$$ for solving this question. The given answer is \begin{aligned}ax^2 + by^2 = C\end{aligned}


3 Answers 3


A start: Let $u=yy'$. Then $x\frac{du}{dx}+u=0$.

I expect you can finish from here.

  • $\begingroup$ Clever substitution. (+1) $\endgroup$
    – user71352
    Aug 9, 2014 at 5:54
  • $\begingroup$ the solution comes out as\begin{aligned}u=\frac{c_1}{x}\end{aligned} \begin{aligned}\implies yy'=\frac{c_1}{x}\end{aligned} \begin{aligned}\implies ydy=\frac{c_1}{x}dx\end{aligned} on integrating further we get, \begin{aligned}\implies \frac{y^2}{2}=c_1lnx+C\end{aligned} how does it take the exact form of answer that is given? $\endgroup$
    – Ritu
    Aug 9, 2014 at 5:58
  • $\begingroup$ $\log|u| = -\log|x| + C \implies |u| = \frac{K}{|x|}$, not $K|x|$. $\endgroup$ Aug 10, 2014 at 1:10

Note that:

Given $$y^2$$

If we take the first derivative by implicit differentiation we get


Again taking a derivative we get

$$2(y')^2 + 2y''$$

This leads us to guess $u = y^2$ is a wise substitution.

From it the equation becomes

$$\frac{1}{2}(xu'') + \frac{1}{2}u' = 0$$

Now we divide out the 1/2 and make another substitution $u' = w$ to obtain:

$$xw' + w = 0$$

You can take it from here, once solved go backwards through the chain of substitutions


I took a different approach and got the same answer as those given above. Since $$ x(yy''+y'^2)+yy'=0 $$is equivalent to $$ x\frac{d}{dx}(yy')=-yy' $$or $$ \frac{\frac{d}{dx}(yy')}{yy'}=-\frac{1}{x} $$then $$ \frac{d}{dx}ln(yy')=-\frac{1}{x} $$This integrates to give $$ ln(yy')=ln(x^{-1})+lnC_1 $$which simplifies to $$ yy'=\frac{C_1}{x} $$From here I used a Lie group $G(x,y)=(\lambda x,\lambda^\beta y), \lambda_o=1$ with stabilizers $\mu=\frac{y}{x^\beta}$ and $\nu=\frac{y'}{x^{\beta -1}}$ to rewrite the equation.
$$ \lambda^\beta y \lambda^{\beta -1} y'=\frac{C_1}{\lambda x} $$For the $\lambda$ terms to cancel out, $\beta =0$. Therefore, the first two nontrivial group stabilizers are $\mu=y$ and $\nu=xy'$. Rewriting the ODE in terms of its stabilizers, $$ \nu=\frac{C_1}{\mu} $$Since $$ x\frac{d\mu}{dx}=\nu=\frac{C_1}{\mu} $$we have $$ \frac{\mu ^2}{2}=C_1ln(x)+C_2 $$Allowing the constants to absorb the factor of two and remembering that $\mu=y$, $$ y=\sqrt{C_1ln(x)+C_2} $$There's nothing original about this solution, but I thought you might enjoy seeing an alternate approach to the problem itself. I picked up the Lie theory in Oak Ridge, and since so few people have ever seen it I like to share it when I can.


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